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Class : IX
CONTENTS
S.No.
1.
Topics
Number System
Page No.
1 - 35
2.
Polynomials
36 - 56
3.
Coordinate Geometry
57 - 62
4.
Linear Equation in two Variable
63 - 71
5.
Introduction of Euclid’s Geometry
72 - 77
6.
Lines and Angles
78 - 88
7.
Triangles
89 - 98
8.
Quadrilateral
99 - 114
9.
Area of parallelograms and triangle
115 - 128
10.
Circle
129 - 155
11.
Constructions
156 - 164
12.
Heron’s Formula
165 - 172
13.
Surface Area & Volume
173 - 181
14.
Statistics
182 - 197
15.
Probability
198 - 204
16.
Proof In Mathematics
205 - 211
17.
Mathematical Modeling
212 - 220
NUMBER SYSTEM 
CLASSIFICATION OF NUMBERS
(I) Natural numbers:
Set of all non-fractional number from 1 to + ∞ , N = {1,2,3,4,....}.
(II) Whole numbers :
Set of numbers from 0 to + ∞ , W = {0,1,2,3,4,.....}.
(III) Integers :
Set of all-non fractional numbers from ∞ to + ∞ , I or Z = (...., -3,-2,-1,0,1,2,3,....}.
(IV) Rational numbers :
These are real numbers which can be expressed in the form of p/q, where p and q are integers and q ≠ 0.
e.g. 2/3, 37/15, -17/19.
All natural numbers, whole numbers and integers are rational.
Rational numbers include all Integers (without any decimal part to it), terminating fractions (fractions in
which the decimal parts terminating e.g. 0.75, - 0.02 etc.) and also non-terminating but recurring decimals
e.g. 0.666...., -2.333....., etc.
Fractions :
Free
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(a) Common fraction
: Fractions whose denominator is not 10.
(b) Decimal fraction
: Fractions whose denominator is 10 or any power of 10.
3
: Numerator < Denominator i.e. .
5
5
: Numerator > Denominator i.e. .
3
2
: Consists of integral as well as fractional part i.e. 3 .
7
(c) Proper fraction
(d) Improper fraction
(e) Mixed fraction
(f) Compound fraction : Fraction whose numerator and denominator themselves are fractions. i.e.
2 /3
.
5 /7
Improper fraction can be written in the form of mixed fractions.
(v) Irrational Numbers :
All real number which are not rational are irrational numbers. These are non-recurring as well as nonterminating type of decimal numbers e.g.
2 , 3 4 ,2 + 3 , 2 + 3 , 4
7
3 etc.
(vi) Real numbers : Number which can represent actual physical quantities in a meaningful way are known
as real numbers. These can be represented on the number line. Number line in geometrical straight line
with arbitrarily defined zero (origin).
(vii) Prime number : All natural numbers that have one and itself only as their factors are called prime
numbers i.e. prime numbers are exactly divisible by 1 and themselves. e.g. 2,3,5,7,11,13,17,19,23....etc. If P is
the set of prime number then P = {2,3,5,7....}.
(viii) Composite numbers : All natural number, which are not prime are composite numbers. If C is the set
of composite number then C = {4,6,8,9,10,12,.....}.
1 is neither prime nor composite number.
(ix) Co-prime numbers : If the H.C.F. of the given numbers (not necessarily prime) is 1 then they are known
as co-prime numbers. e.g. 4, 9, are co-prime as H.C.F. of (4, 9) = 1.
Any two consecutive numbers will always be co-prime.
(x) Even Numbers : All integers which are divisible by 2 are called even numbers. Even numbers are
denoted by the expression 2n, where n is any integer. So, if E is a set even numbers, then E = {...., -4, -2, 0, 2,
4,....}.
(xi) Odd Numbers: All integers which are not divisible by 2 are called odd numbers. Odd numbers are
denoted by the general expression 2n - 1 where n is any integer. If O is a set of odd numbers, then O = {...., 5, -3, -1, 1, 3, 5,....}.
(xii) Imaginary Numbers: All the numbers whose square is negative are called imaginary numbers. e.g. 3i,
4i, i, ..... where i =
−1 .
(xiii) Complex Numbers : The combined form of real and imaginary numbers is known as complex
numbers. It is denoted by Z = A + iB where A is real part and B is imaginary part of Z and A, B ∈ R.
The set of complex number is the super set of all the sets of numbers.
IDENTIFICATION PRIME NUMBER
Free
Step 1 : Find approximate square root of given number.
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Step 2 : Divide the given number by prime numbers less than approximate square root of number. If given
number is not divisible by any of this prime number then the number is prime otherwise not.
Ex.1
571, is it a prime ?
Sol.
Approximate square root of 571 = 24.
Prime number < 24 are 2, 3, 5, 7, 11, 13, 17, 19, & 23. But 571 is not divisible by any of these prime numbers
so 571 is a prime number.
Ex.2
Is 1 prime or composite number ?
Sol.
1 is neither prime nor composite number.
REPRESENTATIO FO RATIONAL NUMBER OF A REAL NUMBER LINE
(i) 3/7 Divide a unit into 7 equal parts.
(ii)
13
7
(iii) −
4
9
(a) Decimal Number (Terminating) :
(i) 2.5
(ii) 2.65 (process of magnification)
Ex.3
Free
Visualize the representation of 5.37 on the number line upto 5 decimal place. i.e. 5.37777.
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(b) Find Rational Numbers Between Two Integral Numbers :
Ex.4
Find 4 rational numbers between 2 and 3.
Sol.
Steps :
(i) Write 2 and 3 multipling in Nr and Dr with (4+1).
(ii) i.e. 2
2 × ( 4 + 1) 1
3 × ( 4 + 1) 15
= &3 =
=
( 4 + 1)
5
( 4 + 1)
5
(iii) So, the four required numbers are
11 12 13 14
, , , .
5 5 5 5
Ex.5
Find three rational no’s between a and b (a < b).
Sol.
a<b
⇒ a+a<b+a
⇒ 2a < a + b
a+b
⇒ a<
2
Again, a < b
⇒ a + b < b + b.
⇒ a + b < 2b
a+b
⇒
< b.
2
a+b
∴ a<
< b.
2
a+b
i.e.
lies between a and b.
2
Hence 1st rational number between a and b is
a+b
.
2
For next rational number
a+b
2a + a + b
a+
3a + b
3a + b a + b
2 =
2
∴
=
a<
<
< b.
2
2
4
4
2
a+b
+b
a + b + 2 b a + 3b
2
Next,
=
=
2
2×2
4
3a + b a + b a + 3b
∴
a<
<
<
< b , and continues like this.
4
2
4
Ex.6
Find 3 rational numbers between
Sol.
1 1
2+3
+
5
1st Method 3 2 = 6 =
2
2
12
Free
1
1
& .
3
2
∴
1 5 1
, ,
3 12 2
=
1 5
4+5
+
3 12 = 12 = 9
2
2
24
∴
1 9 5 1
, , ,
3 24 12 2
=
5 1
5
6
+
+
12 2 = 12 12 = 11
2
2
24
∴
1 9 5 11 1
, , , , .
3 24 12 24 2
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Verify :
8
9
10 11 12  8
1 1
. as
<
<
<
<
= & 
24 24 24 24 24  24 3 2 
2nd Method : Find n rational numbers between a and b (a < b).
(i) Find d =
Ex.7
b−a
.
n+1
(ii) 1st rational number will be a + d.
2nd rational number will be a + 2d.
3rd rational number will be a + 3d and so on....
nth rational number is a + nd.
3
4
Find 5 rational number between
and
5
5
4 3
−
3
4
b−a 5 5 1 1
1
=
= × =
.
Here, a = , b = d =
5
5
n+1
5+1
5 6 30
1st = a + b =
3 1
19
+
=
,
5 30 20
3rd = a + 3d =
3 3
21
+
=
,
5 30 30
5th = a + 5d =
3 5
23
+
=
.
5 30 30
2nd = a + 2d =
3 2
+
,
5 30
4th = a + 4d =
3 4
22
+
=
,
5 30 30
RATIONAL NUMBER IN DECIMAL REPRESENTATION
(a) Terminating Decimal :
In this a finite number of digit occurs after decimal i.e.
1
= 0.5, 0.6875, 0.15 etc.
2
(b) Non-Terminating and Repeating (Recurring Decimal) :
In this a set of digits or a digit is repeated continuously.
Ex.8
2
= 0.6666 − − − − − −− = 0.6 .
3
Ex.9
5
= 0.454545 − − − − − − = 0.45 .
11
PROPERTIES OF RATIONAL NUMBER
If a,b,c are three rational numbers.
(i) Commutative property of addition. a + b = b + a(ii) Associative property of addition (a+b)+c = a+(b+c)
(iii) Additive inverse a + (-a) = 0 0 is identity element, -a is called additive inverse of a.
(iv) Commutative property of multiplications a.b. = b.a. (v) Associative property of multiplication (a.b).c = a.(b.c)
1
(vi) Multiplicative inverse (a ) ×   = 1
a
Free
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1 is called multiplicative identity and
1
is called multiplicative inverse of a or reciprocal of a.
a
(vii) Distributive property a.(b+c) = a.b + a.c
L.H.S. ≠ R.H.S.
∴
Hence in contradicts our assumption that 2 +
2+
∴
3 rational.
3 is irrational.
Ex.9
Prove that
Sol.
Let
3 − 2 is an irrational number
3 − 2 = r where r be a rational number
Squaring both sides
(
⇒
3− 2
)
2
= r2
⇒ 3 + 2 - 2 6 = r2
⇒ 5 - 2 6 = r2
Here,
5 - 2 6 is an irrational number but r2 is a rational number
L.H.S. ≠ R.H.S.
∴
Hence it contradicts our assumption that
3 − 2 is a rational number.
(b) Irrational Number in Decimal Form :
2 = 1.414213 ...... i.e. it is not-recurring as well as non-terminating.
3 = 1.732050807 ...... i.e. it is non-recurring as well as non-terminating.
Ex.10
Insert an irrational number between 2 and 3.
Sol.
2×3 = 6
Ex.11
Find two irrational number between 2 and 2.5.
Sol.
1st Method :
2 × 2.5 = 5
Since there is no rational number whose square is 5. So
Also
5 is irrational..
2 × 5 is a irrational number.
2nd Method : 2.101001000100001.... is between 2 and 5 and it is non-recurring as well as non-terminating.
Also, 2.201001000100001......... and so on.
Ex.12
Find two irrational number between
Sol.
1st Method :
2× 3 =
6 =46
2 and
3.
Irrational number between 2 and
4
6
2 ×4 6 = 4 2 ×8 6
2nd Method : As
2 = 1.414213562 ...... and 3 = 1.732050808......
As ,
2 has 4 in the 1st place of decimal while
∴
Free
3 > 2 and
3 has 7 is the 1st place of decimal.
1.501001000100001......., 1.601001000100001...... etc. are in between
2 and
3
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Ex.13
Find two irrational number between 0.12 and 0.13.
Sol.
0.1201001000100001......., 0.12101001000100001 .......etc.
Ex.14
Find two irrational number between 0.3030030003..... and 0.3010010001 .......
Sol.
0.302020020002...... 0.302030030003.... etc.
Ex.15
Find two rational number between 0.2323323332..... and 0.252552555255552.......
Sol.
1st place is same 2.
2nd place is 3 & 5.
3rd place is 2 in both.
4th place is 3 & 5.
Let a number = 0.25, it falls between the two irrational number.
Also a number = 0.2525 an so on.
(c) Irrational Number on a Number Line :
Ex.16
Plot
2 , 3 , 5 6 on a number line.
Sol.
Another Method for :
(i) Plot
2, 3
So, OC =
(ii) Plot
2 and OD =
3
5, 6, 7 8
OC =
5
OD =
6
OH =
7 .......
(d) Properties of Irrational Number :
(i) Negative of an irrational number is an irrational number e.g. − 3 − 4 5 are irrational.
(ii) Sum and difference of a rational and an irrational number is always an irrational number.
(iii) Sum and difference of two irrational numbers is either rational or irrational number.
(iv) Product of a non-zero rational number with an irrational number is either rational or irrationals
(v) Product of an irrational with a irrational is not always irrational.
Ex.17
Two number’s are 2 and
Sum = 2 +
3 , is an irrational number.
Difference = 2 Also
Free
3 , then
3 , is an irrational number.
3 − 2 is an irrational number.
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Ex.18
Two number’s are 4 and
Sum = 4 +
3
3
3 , is an irrational number.
Two irrational numbers are
(
Sum =
3 ,− 3 , then
)
3 − (− 3 ) = 2
3 + − 3 = 0 which is rational.
Difference =
Ex.20
3 , then
3 , is an irrational number.
Difference = 4 -
Ex.19
3
3 , which is irrational.
Two irrational numbers are 2 +
(
) (
3 and 2 - 3 , then
)
Sum = 2 + 3 + 2 − 3 = 4 , a rational number
Two irrational numbers are
Difference =
Ex.21
3 + 3 − 3 + 3 = 6 , a rational number
Two irrational numbers are
Sum =
Ex.22
3 + 3m 3 − 3
3 − 2 , 3 + 2 , then
3 − 2 + 3 + 2 = 2 3 , an irrational
2 is a rational number and
3 is an irrational.
2 × 3 = 2 3 , an irrational.
Ex.23
0 a rational and
0×
Ex.24
3 an irrational.
3 = 0, a rational.
4
4
4
× 3=
3=
is an irrational.
3
3
3
Ex.25
3 × 3 = 3 × 3 = 9 = 3 a rational number.
Ex.26
2 3 × 3 2 = 2 × 3 3 × 2 = 6 6 and irrational number.
Ex.27
3
Ex.28
(2 + 3 )(2 − 3 ) = (2) − ( 3 )
Free
3 × 3 3 2 = 3 3 × 3 2 = 3 3 3 = 3 a rational number.
2
2
= 4 − 3 = 1 a rational number.
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Ex.29
(2 + 3 )(2 + 3 ) = (2 + 3 )
= (2) + ( 3 )
2
2
2
+ 2( 2 ) ×
( 3)
= 4+3+4 3
= 7 + 4 3 an irrational number
NOTE :
(i)
− 2 ≠ − 2 , it is not a irrational number.
(ii)
−2× −3 ≠
Instead
(
− 2 × −3 = 6
)
− 2 , − 3 are called Imaginary numbers.
− 2 = i 2 , where i ( = iota) =
∴
−1
(A) i2 = -1
(B) i3 = i2 × i = (-1) × i = -i
(C) i4 = i2 × i2 = (-1) × (-1) = 1
(iii) Numbers of the type (a + ib) are called complex numbers where (a, b) ∈ R.e.g. 2 + 3i, -2 + 4i, -3i, 11 - 4i,
are complex numbers.
GEOMETRICAL REPRESENTATION OF REAL NUMBERS
To represent any real number of number line we follows the following steps :
STEP I : Obtain the positive real number x (say).
STEP II : Draw a line and mark a point A on it.
STEP III : Mark a point B on the line such that AB = x units.
STEP IV : From point B mark a distance of 1 unit and mark the new point as C.
STEP V : Find the mid - point of AC and mark the point as O.
STEP VI : Draw a circle with centre O and radius OC.
STEP VII : Draw a line perpendicular to AC passing through B and intersecting the semi circle at D.
Length BD is equal to
Free
x.
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4.
Examine whether the following numbers are rational or irrational :
(
(i) 2 − 3
)
2
(ii)
(
2+ 3
(
)
2
5.
Represent
6.
Represent ( 2 + 3 ) on the number line.
7.
Prove that
8.
Prove that
9.
Prove that ( 2 + 2 ) is an irrational number.
10.
Multiply
11.
Express the following in the form of p/q.
)
(iv)
3 −1
3 +1
8.3 on the number line.
(
)
2 + 5 is an irrational number.
7 is not a rational number.
27 a 3b 2 c 4 × 3 128a7 b9 c 2 × 6 729ab12 c2 .
(i) 0.3
12.
)(
(iii) 3 + 2 3 − 2
(ii) 0.37
(iii) 0.54
(iv) 0.05
(v) 1.3
(vi) 0.621
Simplify : 0.4 + .018
NUMBER SYSTEM

SURDS
Any irrational number of the form
n
a is given a special name surd. Where ‘a’ is called radicand, it should
always be a rational number. Also the symbol
n
is called the radical sign and the index n is called order
of the surd.
1
n
a is read as ‘nth root a’ and can also be written as a n .
(a) Some Identical Surds :
(i)
3
4 is a surd as radicand is a rational number.
Similar examples
3
5 , 4 12 , 5 7 , 12 ,.........
(i) 2 3 is a surd (as surd + rational number will give a surd)
Similar examples
Free
3 + 1, 3 3 + 1,....
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(
7 − 4 3 is a surd as 7 - 4 3 is a perfect square of 2 − 3
(iii)
Similar examples
)
7 + 4 3 , 9 − 4 5 , 9 + 4 5 ,.........
1
(i)
3
3 is a surd as
Similar examples
3 3
3
1
 1 3
3 = 32  = 36 = 6 3
 
 
5 , 4 5 6 ,....... ...
(b) Some Expression are not Surds :
(i)
3
8 because
8 = 3 2 3 = 2 , which is a rational number.
2 + 3 because 2 + 3 is not a perfect square.
(ii)
(iii)
3
3
1 + 3 because radicand is an irrational number.
LAWS OF SURDS
(i)
( a)
n
e.g. (A)
(ii)
n
3
= n an = a
3
8 = 3 23 = 2
(B)
4
81 = 4 3 4 = 3
a × n b = n ab
e.g. (A)
but,
n
3
[Here order should be same]
2 × 3 6 = 3 2 × 6 = 3 12
3 × 4 6 ≠ 3×6
[Because order is not same]
1st make their order same and then you can multiply.
Free
a +nb =n
a
b
(iii)
n
(iv)
nm
(v)
n
a=
or,
n
am =
e.g.
3
6 2 make its order 6, then
3
6 2 = 3× 2 6 2× 2 = 6 6 4 .
e.g.
3
6 make its order 15, then
3
6 = 3×5 61×5 = 15 6 5 .
a =
nm
n×p
a =mna
e.g. =
ap
n ×p
2 =88
[Important for changing order of surds]
a m ×p
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OPERATION OF SURDS
(a) Addition and Subtraction of Surds :
Addition and subtraction of surds are possible only when order and radicand are same i.e. only for surds.
Ex.1
Simplify
(i)
6 − 216 + 96 = 15 6 − 6 2 × 6 + 16 × 6
[Bring surd in simples form]
= 15 6 − 6 6 + 4 6
= (15 - 6 + 4 ) 6
Ans.
= 13 6
(ii) 53 250 + 7 3 16 − 143 54
= 53 125 × 2 + 7 3 8 × 2 − 143 27 × 2
= 5 × 53 2 + 7 × 2 3 2 − 14 × 3 × 3 2
= ( 25 + 14 − 42 )3 2
= −33 2
(ii) 53 250 + 7 3 16 − 143 54
Ans.
= 53 125 × 2 + 7 3 8 × 2 − 143 27 × 2
= 5 × 53 2 + 7 × 2 3 2 − 14 × 3 × 3 2
= ( 25 + 14 − 42 )3 2
= −33 2
(iii) 4 3 + 3 48 −
Ans.
5 1
5 1× 3
= 4 3 + 3 16 × 3 −
2 3
2 3× 3
= 4 3 + 3× 4 3 −
= 4 3 + 12 3 −
5 1
×
3
2 3
5
3
6
5

=  4 + 12 −  3
6

=
91
3
6
Ans.
(b) Multiplication and Division of Surds :
Ex.2
3
(i)
3
4 × 3 22 = 3 4 × 22 = 2 3 × 11 = 2 3 11
(i)
3
2 ×4 3 =
Ex.3
Simplify
Hint :
6
Free
12
2 4 × 12 3 3 = 12 2 4 × 3 3 = 12 16 × 27 =
12
432
8a 5b × 3 4a 2 b 2
83 a15 b 3 × 6 4 2 a 4 b 4 = 6 2 13 a19 b7 = 6 2ab .
Ans.
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Ex.4
24 ÷ 3 200 =
Divide
24
=
200
3
( 24)3
6
6
( 200)
=6
2
216
625
Ans..
(c) Comparison of Surds :
It is clear that if x > y > 0 and n > 1 is a positive integer then
n
x >ny.
16 > 3 12 , 5 35 > 5 25 and so on.
Ex.5
3
Ex.6
Which is greater is each of the following :
(i)
3
16 and
5
1
and
2
(ii)
8
L.C.M. of 3 and 5 15.
1
3
3
L.C.M. of 2 and 3 is 6.
6 = 3× 5 6 5 = 15 7776
6
1
1
  and 3  
2
3
2
3
3
5
8 = 3×5 85 = 15 512
6
1
1
and 6
8
9
1 1

As 8 < 9 ∴ 8 > 9 
∴
75
7776 > 15 512
so,
⇒
3
6 >58
⇒
2 , 3 3 and
Ex.7
Arrange
Sol.
L.C.M. of 2, 3, 4 is 12.
4
6
1 6 1
>
8
9
1
>
2
3
1
3
5 is ascending order.
2 = 2×6 2 6 = 12 64
∴
3
3=
3× 4
3 4 = 12 81
4
5=
4× 3
53 = 12 125
As, 64 < 81 < 125.
∴
12
⇒
Ex.8
Sol.
64 < 12 81 < 12 125
2 <33<45
Which is greater
7− 3=
And,
7 − 3 or
( 7 − 3 )( 7 + 3 )
=
( 7 + 3)
5 −1 =
Free
7−3
=
7+ 3
( 5 − 1)( 5 + 1)
5−1
=
=
( 5 + 1)
5 +1
Now, we know that
So,
5 − 1?
7 > 5 and
4
7+ 3
4
5 +1
3 > 1 , add
7 + 3 > 5 +1
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⇒
1
<
7+ 3
1
5 +1
So,
5 −1 > 7 − 3
4
<
7+ 3
⇒
4
5 +1
⇒
7 − 3 < 5 −1
RATIONALIZATION OF SURDS
Rationalizing factor product of two surds is a rational number then each of them is called the rationalizing
factor (R.F.) of the other. The process of converting a surd to a rational number by using an appropriate
multiplier is known as rationalization.
Some examples :
(i) R.F. of
a is
)
a× a =a .
a is 3 a 2 ∴ 3 a × 3 a 2 = 3 a 3 = a  .


3
(ii) R.F. of
(∴
a
[ ( a + b )( a − b ) = a − b].
b & vice versa [∴ (a + b )(a − b ) = a − b]
a − ab + b  ∴ ( a + b ) a − ab + b 



a + b is a − b & vice versa ∴
(iii) R.F. of
2
(iv) R.F. of a + b is a −
3
(v) R.F. of
∴

3
3
(
) (
a + b + c is
10 (ii)
(i)
(iii)
12
(viii) 7 − 4 3
2
3
3
3
2
3
2
(vi)
4
3
) (
)
a + b − c nd a + b − c + 2 ab .
(ix) 3 3 + 2 2 (x)
[∴ 10 × 10 = 10 × 10 = 10]
(ii).
3
3
4 (v)
3 +3 2
3
16
162
(vii) 2 + 3
(xi) 1 + 2 + 3
First write it’s simplest from i.e. 2 3 .
Now find R.F. (i.e. R.F. of
∴ R.F. of
3 is
3)
3 Ans.
12 is
162 Sol.
Simplest from of
∴ R.F. of
3
3
4 × 42 = 43 = 4
3
4 Sol.
(v).
3
16
Sol.
Simplest from of
∴ R.F. of
3
3
16 is
162 is 9 2 . R.F. of
2 is
2.
2 Ans.
162 is
(iv)
3
as 10 is rational number.
10 Ans.
10 is
12 Sol.
(iii)
(iv)
162
10
∴ R.F. of
Free
3
3
Find the R.G. (rationalizing factor) of the following :
(i)
Sol.
3
3
2
= a + b which is rational.

( a) + ( b)
(vi) R.F. of
Ex.9
3
a + 3 b is 

∴
R.F. of
16 is 2 3 2
Now R.F. of
3
Ans.
22
3
3
4 is
2 is
3
3
42
Ans.
22
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(vi)
4
Sol.
Simplest form of
162
4
Now R.F. of
4
162 is 3 4 2
2 is 4 2 3
R.F. of ( 4 162 ) is
4
23
Ans.
(vii)
2+ 3
Sol.
As 2 + 3 2 − 3 = ( 2 )2 −
(
)(
)
( 3)
2
= 4 − 3 = 1, , which is rational.
∴ R.F. of ( 2 + 3 ) is ( 2 − 3 ) Ans.
(viii)
7−4 3
Sol.
As 7 − 4 3 7 + 4 3 = (7 )2 − 4 − 3
(
)(
)
(
(
∴ R.F. of (7 − 4 3 ) is 7 + 4 3
3 3 +2 2
Sol.
As 3 3 + 2 2 3 3 − 2 2 = 3 3
)(
2
= 49 - 48 = 1, which is rational
)
(ix).
(
)
Ans.
) ( ) − (2 2 )
2
2
= 27 − 8 = 19 , which is rational.
∴ R.F. of ( 3 3 + 2 2 ) is ( 3 3 − 2 2 ) Ans.
3 +3 2
(x)
3
Sol.
As
(
3
3
3
3 + 3 2  3 32 − 3 3 × 3 2 + 2 2  =  3 33 + 2 3  = 3 + 2 = 5, which is rational.

 

)
∴ R.F. of ( 3 3 + 3 2 ) is  3 32 − 3 3 × 3 2 + 3 2 2  Ans.


(xi)
1+ 2 + 3
Sol.
(1 +
)(
) (
) − ( 3)
+ ( 2 ) + 2( 1)( 2 ) − 3
2 + 3 1+ 2 − 3 = 1+ 2
= 1)2
2
2
2
= 1+2+2 2 −3
= 3+2 2 −3
=2 2
2 2 × 2 = 2×2 = 4
(
)
∴ R.F. of 1 + 2 + 3 is 1 + 2 − 3 and
NOTE : R.F. of
Free
a + b or
2 . Ans.
a − b type surds are also called conjugate surds & vice versa.
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Ex.10
(i)
(ii)
2 − 3 is conjugate of 2 + 3
5 + 1 is conjugate of
5 −1
NOTE : Sometimes conjugate surds and reciprocals are same.
Ex.11
(i) 2 + 3 , it’s conjugate is 2 − 3 , its reciprocal is 2 − 3 & vice versa.
(ii) 5 − 2 6 , it’s conjugate is 5 + 2 6 , its reciprocal is 5 − 2 6 & vice versa.
(iii) 6 − 35 ,6 + 35
(iv) 7 − 4 3 , 7 + 4 3
............... and so on.
(v) 8 + 3 7 , 8 − 3 7
Ex.12
Express the following surd with a rational denominator.
 ( 15 + 1) + (
8
[( 15 + 1) − ( 15 + 3 )] ×  ( 15 + 1) + (
8( 15 + 1 + 5 + 3 )
=
( 15 + 1) − ( 5 + 3 )
8( 15 + 1 + 5 + 3 )
=
15 + 1 + 2 15 − (5 + 3 + 2 15 )
8( 15 + 1 + 5 + 3 )
=
8
=
15 + 1 − 5 − 3
Sol.
2
)
)
5+ 3 

5 + 3 
2
8
= ( 15 + 1 + 5 + 3 )
Ex.13
Sol.
Rationalize the denominator of
a2
a 2 + b2 + b
=
a2
a 2 + b2 + b
×
Ans.
a2
a2 + b2 + b
a2 + b2 − b
a2 + b2 − b
a 2  a 2 + b 2 − b 


=
2
 a 2 + b 2  − ( b)2


a 2  a 2 + b 2 − b 
 =  a2 + b2 − b 
= 2




a + b2 − b 2
Free
Ans.
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3+2 2
= a + b 2 , where a and b are rational then find the values of a and b.
3− 2
Ex.14
If
Sol.
L.H.S.
∴
3 + 2 2 ( 3 + 2 2 )( 3 + 2 )
=
3− 2
( 3 − 2 )(3 + 2 )
=
9+3 2 +6 2 +4
9−2
=
13 + 9 2
7
=
13 9
+
2
7 7
13 9
+
2 = a+b 2
7 7
Equating the rational and irrational parts
We get a =
Ex.15
If
Ans.
3 = 1.732 , find the value of
1
=
3 −1
Sol.
13
9
,b =
7
7
1
3 −1
1
3 +1
×
3 −1
3 +1
=
3 +1
3−1
=
3 +1
2
=
1.732 + 1
2
=
2.732
2
Ans.
= 1.366
Ex.16
If
5 = 2.236 and
Evaluate :
Sol.
Free
3
+
5+ 2
2 = 1.414, then
3
+
5+ 2
4
5− 2
4
3 5 − 2 + 4( 5 + 2 )
=
5− 2
( 5 − 2 )( 5 + 2 )
=
3 5 −3 2 +4 5 +4 2
5−2
=
7 5+ 2
5−2
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=
7 5+ 2
3
=
7 × 2.236 + 1.414
3
=
15.652 + 1.414
3
=
17.066
3
= 5.689 (approximate)
1
find the value of x3 - x2 - 11x + 3.
2+ 3
Ex.17
If =
Sol.
As, x =
1
=2− 3
2+ 3
⇒ x-2=-
3
(
⇒ (x - 2)2 = − 3
)
2
[By squaring both sides]
⇒ x2 + 4 - 4x = 3
⇒ x2 - 4x + 1 = 0
Now,
x3 - x2 - 11x + 3 = x3 - 4x2 + x + 3x2 - 12x + 3
= x (x2 - 4x + 1) + 3 (x2 - 4x + 1)
= x(0) + 3 (0)
Ans.
=0+0=0
Ex.18
If x = 3 - 8 , find the value of x 3 +
Sol.
x=3-
1
.
x3
8
∴
1
1
=
x 3− 8
⇒
1
= 3+ 8
x
Now,
x+
1
= 3− 8 +3+ 8 = 6
x
3
Free
⇒
x3 +
1 
1
1
1
=  x +  − 3x  x + 
3
x
x
x
x

⇒
x3 +
1
= (6 )3 − 3( 6)
x3
⇒
x3 +
1
= 216 − 18
x3
⇒
x3 +
1
= 198
x3
Ans.
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Ex.19
Sol.
If x = 1 + 21/3 + 2 2/3, show that x3 - 3x2 - 3x - 1 = 0
⇒
⇒
⇒
⇒
⇒
⇒
⇒
Ex.20
Sol.
x = 1 + 21/3 + 2 2/3
x - 1 (2 1/3 + 22/3)
(x - 1)3 = (2 1/3 + 22/3)3
(x - 1)3 = (21/3) + (2 2/3)3 + 3.21/3.2 2/3(2 1/3 + 2 1/3-)
(x - 1)3 = 2 + 22 + 3.21 (x - 1)
(x - 1)3 = 6 + 6 (x - 1)
x3 - 3x2 + 3x - 1 = 6x
x3 - 3x2 - 3x - 1 = 0
Ans.
Solve :
x + 3 + x − 2 = 5.
x+3 = 5− x−2
Ex.21
⇒
(
⇒
⇒
⇒
⇒
⇒
⇒
⇒
x + 3 = 25 + (x - 2) - 10 x − 2
x + 3 = 25 + x - 2 - 10 x − 2
3 - 23 = - 10 x − 2
-20 = -10 x − 2
2 = x−2
x-2=4
[By squaring both sides]
x=6
Ans.
x+3
If x = 1 +
) = (5 −
2
x−2
)
2
[By squaring both sides]
2 + 3 , prove that x4 - 4x3 - 4x2 + 16 - 8 = 0.
Hint : x = 1 + 2 + 3
⇒ x-1= 2 + 3
(
)
2
⇒
( x − 1)2 =
⇒
⇒
⇒
⇒
⇒
⇒
x2 + 1 - 2x = 2 + 3 + 2 6
x2 - 2x - 4 = 2 6
(x2 - 2x - 4)2 = ( 2 6 )2
x4 + 4x2 + 16 - 4x3 + 16x - 8x2 = 24
x4 - 4x3 - 4x2 + 16x + 16 - 24 = 0
x4 - 4x3 - 4x2 + 16x - 8 = 0
2+ 3
[By squaring both sides]
Ans.
EXPONENTS OF REAL NUMBER
(a) Positive Integral Power :
For any real number a and a positive integer ‘n’ we define an as :
an = a × a × a × ............... x a (n times)
an is called then nth power of a. The real number ‘a’ is called the base and ‘n’ is called the exponent of the nth
power of a.
e.g. 2 3 = 2 × 2 × 2 = 8
NOTE : For any non-zero real number ‘a’ we define a0 = 1.
0
3
e.g. thus, 3 0 = 1, 50,   = 1 and so on.
4
Free
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(b) Negative Integral Power :
For any non-zero real number ‘a’ and a positive integer ‘n’ we define a − n =
1
an
Thus we have defined an find all integral values of n, positive, zero or negative. an is called the nth power of
a.
RATIONAL EXPONENTS OR A REAL NUMBER
(a) Principal of nth Root of a Positive Real Numbers :
If ‘a’ is a positive real number and ’n’ is a positive integer, then the principal nth root of a is the unique
positive real number x such that xn = a.
The principal nth root of a positive real number a is denoted by a1/n or
n
a.
(b) Principal of nth Root of a Negative Real Numbers :
If ‘a’ is a negative real number and ‘n’ is an odd positive integer, then the principle nth root of a is define as |a|1/n i.e. the principal nth root of -a is negative of the principal nth root of |a|.
Remark :
It ‘a’ is negative real number and ‘n’ is an even positive integer, then the principle nth root of a is not
defined, because an even power of real number is always positive. Therefore (-9)1/2 is a meaningless
quantity, if we confine ourselves to the set of real number, only.
(c) Rational Power (Exponents) :
For any positive real number ‘a’ and a rational number
p
≠ where q ≠ 0 , we define ap / q = (a p )1 / q i.e. ap/q
q
is the principle qth root of ap.
LAWS OF RATIONAL EXPONETNS
The following laws hold the rational exponents
(i) a m × an = am+1
(ii) am ÷ an = am-n
(ii) (am)n = amn
(iv) a − n =
(v) am/n = (am)1/n = (a1/n)m i.e. am/n =
a
(vii)  
b
m
=
n
am =
am
bm
( a)
n
m
1
an
(vi) (ab)m = ambm
(viii) abn = a b+b+b….n tmes
Where a,b are positive real number and m,n are relational numbers.
ILLUSTRATIONS :
Ex.22
Evaluate each of the following:
(i) 5 2 × 54
Free
(ii) 58 ÷ 53
( )
(iii) 32
2
 11 
(iv)  
 12 
3
3
(v)  
4
−3
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Sol.
Using the laws of indices, we have
∵a m × a n = a m + n
(i) 5 2.54 = 5 2+4 = 56 = 15625
58
= 58 − 3 = 5 5 = 3125
53
(ii) 58 ÷ 5 3 =
( )
(iii) 32
3
∵a m + a n = am − n
= 32×3 = 36 = 729
∵(am )n = am× n
3
a
∵ 
b
113 1331
 11 
(iv)   = 3 =
1728
12
 12 
3
(v)  
4
Ex.23
−3
=
1
3
 
4
=
3
1
1
64
=
=
27 27
33
64
43
=
∵a − n =
am
bm
1
an
Evaluate each of the following :
4
2
 2   11   3 
(i)   ×   ×  
 11   3   2 
3
5
4
1 −2 3
(ii)   × 
 × 
2  3  5
3
−3
3
2 2
(iv)   ×   ×  
3
5
5
   
(iii) 2 55 × 2 60 − 2 97 × 2 18
Sol.
m
−1
2
We have.
4
2
3
2 4 112 33
 2   11   3 
(i)   ×   ×   = 4 × 2 × 3
11
3
2
 11   3   2 
=
2×3
112
=
6
121
Ans.
(ii) We have,
5
4
1 −2 3
  ×
 × 
2  3  5
Free
−1
1
5
4 

1 −2  3 
=   ×
 ×
2  3   5 
 
 
=
15 (− 2 )2 5
× 4 ×
3
25
3
=
1 × 16 × 5
32 × 81 × 3
=
5
2 × 81 × 3
=
5
486
Ans.
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(iii) We have,
2 55 × 2 60 − 2 97 × 2 18 = 55 + 60 −2 97 + 18
= 2 15 − 2 115
Ans.
=0
(iv) We have,
3
−3
2 2
3
  ×  × 
3
5
   
5
Ex.24
23
1
32
×
× 2
3
(2 / 5) 5
3
=
23
1
32
×
×
3 3 2 3 / 5 3 52
=
2 3 × 53 × 32
3 3 × 2 3 × 52
=
5
3
Ans.
(25)3 / 2 × (243)3 / 5
(16 )5 / 4 × (8)4 / 3
(ii)
16 × 2 n + 1 − 4 × 2 n
16 × 2 n + 2 − 2 × 2 n + 2
We have,
(i)
(ii)
Free
=
Simplify :
(i)
Sol.
−2
(25 )3 / 2 × (243 )3 / 5
(16 )5 / 4 × (8)4 / 3
16 × 2 n + 1 − 4 × 2 n
16 × 2 n + 2 − 2 × 2 n + 2
=
(5 )
(2 )
=
52 ×3 / 2 × 35× 3 / 5
2 4× 5 / 4 × 2 3× 4 / 3
=
53 × 33
25 × 24
=
125 × 27
32 × 16
=
3375
512
=
24 × 2n +1 − 22 × 2n
24 × 2n+2 − 2 × 2n+2
=
2n+5 − 2n+2
2n +6 − 2n +3
=
2n +5 − 2n +2
2.2 n + 5 − 2.2 n + 2
=
2n+5 − 2n+2
1
=
2
2 2n +5 − 2n +2
2 3/2
4 5 /4
(
( )
× (2 )
× 35
3/5
3 4 /3
Ans.
)
Ans.
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Ex.25
 81 
Simplify  
 16 
Sol.
We have
 81 
 
 16 
−3 / 4
−3 / 4
 25 
×  
 9 
−3 / 2
−3
5 
÷  
 2  
 25  − 3 / 2  5  − 3   3 4 
×  
÷    =  4 
 2    2 
 9 
 3  4 
=   
 2  
3
= 
2
−3 / 4
4x −3 / 4
−3 / 4
 5  2 
×   
 3  
−3 / 2
 52
×  2
 3





−3 / 2
−3 
5
÷  
2 

 5  − 3 
÷   
 2  
 5  2 x − 3 / 2  5  −3 
×  
÷  
 2  
 3 
−3 
−3
−3
3
5
5 
=   ×   ÷   
2
 2  
 3 
3
−3
−3
 2   5 
5 
=   ×   ×   
 3   3 
 2  
=
2 3  33 2 3 
× ÷ 
33  53 53 
=
2 3  33 53 
× × 
33  53 2 3 
= 1 Ans.
Free
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EXERCISE
OBJECTIVE DPP - 3.1
1.
If x= 3 + 8 and y = 3 - 8 then
(A) -34
2.
If
(B) 34
5
5 −4
The value of
5
3− 2
If x =
Simplify :
(A) 1
(D) − 6 5
5
6+4
(C)
6 −4
5
(D)
4
(D)
6+4
5
3
3
(C)
3
3
2
(32 )−3 is :
and y =
(C) 1/32
(D) None
3+ 2
the value of x2 + xy + y2 is :
3− 2
(B) 100
(C) 1
(D) 0
(C) 10
(D) 100
(C)
(D)
2
1
3
+
−
5+ 3
3+ 2
5+ 2
(B) 0
Which of the following is smallest ?
(A)
Free
(C) 6 5
(B) 1/16
(A) 99
9.
(B) − 8 5
(B)
2
3+ 2
8.
(D) (8, 3)
(B)
(A) 1/8
7.
(C) (-8, 3)
Which one is greatest in the following :
(A)
6.
(B) (-8, -3)
x−y
3+ 2
and y = 1, the value of
is :
x − 3y
3− 2
If x =
(A)
5.
(D) -12 8
5 −2
5 +2
−
=
5 +2
5 −2
(A) 8 5
4.
(C) 12 8
3+ 7
= a + b 7 then (a,b) =
3− 7
(A) (8, -3)
3.
1
1
+
=
x2 y2
4
5
(B)
5
4
4
3
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10
The product of
6
(A)
11.
3 and
The exponential from of
15.
(C)
675
6
575
(D)
6
475
2 × 2 × 2 is :
(C) 23/4
(D) 81/2
(C) 3
(D) 5
The value of x, if 5 x-3 . 32x-8 = 225, is :
If 25x ÷ 2x =
(B) 2
5
2 20 then x =
(A) 0
14.
6
(B) 83/4
(A) 1
13.
5 is :
(B)
375
(A) 21/16
12.
3
3
1
2
(B) - 1
(C)
(B) 81
(C) 243
(D) 1
(729 )2.5 =
(A)
1
81
4 3
x2 =
(A) x
(D) 729
1
1
1
(B) x 2
(C) x 3
(D) x 6
SUBJECTIVE DPP - 3.2
1.
Arrange the following surds in ascending order of magnitude :
(i) 4 10 ,3 6 , 3
2.
(ii) 3 4 ,4 5 , 3
Whish is greater :
17 − 12 or 11 − 6 .
8
.
15 + 1 − 5 − 3
3.
Simplify :
4.
If p and q are rational number and p − q =
5.
Find the simplest R.F. of :
(i)
6.
32
(ii)
3
36
(iii) 2 3 / 5
Retionalise the denominator :
(i)
Free
3
4+ 2
find p and q.
3+ 2
3
5
(ii)
2+ 5
3
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7.
Retionalise the denominator and simplify :
3− 2
3+ 2
(i)
8.
4 3 +5 2
48 + 18
(iii)
Simplify :
5+ 3
(i)
+
5− 3
9.
1+ 2
3+2 2
(ii)
5− 3
5+ 3
(ii)
7+3 5 7−3 5
−
3+ 5
3− 5
(ii)
5+ 6
=a+b 6
5− 6
Find the value of a and b
11 − 7
= a − b 77
11 + 7
(i)
3 +1
find the value of 4x3 + 2x2 - 8x + 7.
2
10.
If x =
11.
If x =
12.
Show that a = x + 1/x, where x =
13.
Prove that :
14.
If x =
15.
Evaluate:
5 − 21
1  
1  
1

show that  x 3 + 3  − 5 x 2 + 2  +  x +  = 0.
2
x
x  
x  

a+2 + a−2
.
a+2 − a−2
1
1
1
1
1
−
+
−
+
= 5.
3− 8
8− 7
7− 6
6− 5
5 −2
5+ 2
and y =
5− 2
5 +2 +
5 −2
5− 2
find the value of 3x2 + 4xy - 3y2.
5+ 2
− 3−2 2.
5 +1
16.
If x= 3, b = 4 then find the values of :
(i) a b + ba
17.
−2 / 3
y 4 ÷ xy − 1 / 2 .
Simplify :
[
(i) 16 − 1 / 5
]
5/2
[
(ii) [0.001]
]
9n × 32 × 3 −−n / 2
33m × 2 3
−2
− ( 27 )n
If
20.
Find the value of x, if 5 x-3(2x-3) = 625.
=
1
3
1
, then prove than m - n = 1.
27
19.
Free
(iii) ab - ba
Simplify :
( x)
18.
(ii) a a + bb
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ANSWER KEY
(Objective DPP # 1.1)
Qus.
1
2
3
4
5
6
Ans.
A
B
C
B
D
C
(Subjective DPP # 1.2)
1.
(i) Non-terminating and repeating
(ii) Non-terminating and non-repeating
(iii) Non-terminating and repeating
(iv) Terminating
2.
−7 −4 −3 −5 −11
, , , ,
6 3 2 3 6
3.
-4, -3, -2, -1
4.
−5
24
6.
9 10 11
, ,
24 24 24
7.
−5 −4 −3
,
,
14 14 14
22 23 24 25 26 27
, , , , ,
7 7 7 7 7 7
5.
(Objective DPP # 2.1)
Qus.
1
2
3
4
5
6
7
8
Ans.
A
B
A
A
C
D
C
C
(Subjective DPP # 2.2)
3.
0.110101001000100001
4.
(i) irrational
10.
36A 4B 6C 3 6 108
11.
(i) 1/3
12.
19
30
Free
(ii) irrational
(ii)
37
99
(iii)
(iii) rational
6
11
(iv)
5
99
(iv) irrational
(v)
4
3
(vi)
4
3
(v)
23
37
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(Objective DPP # 3.1)
Qus.
1
2
3
4
5
6
7
8
9
10
Ans.
B
D
B
D
C
A
A
B
B
B
Qus.
11
12
13
14
15
Ans.
C
D
D
C
D
(Subjective DPP # 3.2)
1.
(i) 4 10 > 3 6 > 3
3.
16 + 1 + 5 + 5
3
5 (ii)
2
6.
(i)
8.
(i) 8 (ii)
10.
15.
17.
Free
6 + 15
3
(ii) 4 5 > 3 4 > 3
4. P =
10
2
,Q =
7
49
2.
5. (i)
11 − 6
3
(ii)
2
3
6
9+4 6
15
7.
(i) 5 − 2 6
9.
(i) a = 9/2, b = 1/2
10
14.
12 + 56 10
3
1
16.
(i) 145
(ii) 283
(iii) 17
18.
(i) 1/4
(ii) 0.1
20. 1
y9 / 4
x5 / 6
5
(ii) 7 + 5 2
(iii)
(ii) a =
(iii) 22/5
31
10
,b =
19
19
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POLYNOMIALS
POLYNOMIALS

An algebraic expression (f(x) of the form f(x) = a 0 + a1x + a2x2 + ........ + anxn, where a0,a 1,a2 ......., an are real
numbers and all the index of ‘x’ are non-negative integers is called a polynomials in x.
(a) Degree of the Polynomial :
Highest Index of x in algebraic expression is called the degree of the polynomial, here a0, a 1x , a2x2 ..... anxn,
are called the terms o the polynomial and z 0. a 1, a2......, an are called various coefficients of the polynomial
f(x).
NOTE : A polynomial in x is said to be in standard form when the terms are written either in increasing
order or decreasing order of the indices of x in various terms.
(b) Different Types of Polynomials :
Generally, we divide the polynomials in the following categories.
(i) Based on degrees :
There are four types of polynomials based on degrees. These are listed below :
(A) Linear Polynomials : A polynomials of degree one is called a linear polynomial. The general
formula of linear polynomial is ax + b, where a and b are any real constant and a ≠ 0.
(B) Quadratic Polynomials : A polynomial of degree two is called a quadratic polynomial. The general
form of a quadratic polynomial is ax2 + b + c, where a ≠ 0.
(C) Cubic Polynomials : A polynomial of degree three is called a cubic polynomial. The general form
of a cubic polynomial is ax3 + bx2 + cx + d, where a ≠ 0.
(D) Biquadratic (or quadric) Polynomials : A polynomial of degree four is called a biquadratic
(quadratic) polynomial. The general form of a biquadratic polynomial is ax4 + bx3 + cx2 + dx + e ,
where a ≠ 0.
NOTE : A polynomial of degree five or more than five does not have any particular name. Such a
polynomial usually called a polynomial of degree five or six or ....etc.
(ii) Based on number of terms
There are three types of polynomials based on number of terms. These are as follows :
(A) Monomial : A polynomial is said to be monomial if it has only one term. e.g. x, 9x2, 5x3 all are
monomials.
(B) Binomial : A polynomial is said to be binomial if it contains two terms e.g. 2x2 + 3x,
3 x + 5x3 , -8x3 +
3, all are binomials.
Free
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(C) Trinomials : A polynomial is said to be a trinomial it if contains three terms. e.g. 3x3 - 8 +
5
,
2
7x10 8x4 - 3x2, 5 - 7x + 8x9, are all trinomials.
NOTE : A polynomial having four or more than four terms does not have particular Name. These are
simply called polynomials.
(iii) Zero degree polynomial : Any non-zero number (constant) is regarded as polynomial of degree zero or
zero degree polynomial. i.e. f(x) = a, where a ≠ 0 is a zero degree polynomial, since we can write f(x) = a as
f(x) = ax0.
(iv) Zero polynomial : A polynomial whose all coefficients are zeros is called as zero polynomial i.e. f(x) =
0, we cannot determine the degree of zero polynomial.
ALGEBRAIC IDENTITY
An identity is an equality which is true for all values of the variables
Some important identities are
(i) (a + b)2 = a2 + 2ab + b2
(ii) (a - b)2 = a2 - 2ab + b2
(iii) a 2 - b2 = (a + b) (a - b)
(iv) a 3 + b3 = (a + b) (a2 - ab + b2)
(v) a 3 - b3 = (a - b) (a2 + ab + b2)
(vi) (a + b)3 = a3 + b3 + 3ab (a + b)
(vii) (a - b)3 = a 3 - b3 - 3ab (a - b)
(viii) a 4 + a2b2 + b4 = (a2 + ab + b2) (a2 - ab + b2)
(ix) a3 + b3 + c3 - 3abc = (a + b + c) (a2 + b2 + c2 - ab - bc - ac)
Special case : if a + b + c = 0 then a3 + b3 + c3 = 3abc.
(a) Value Form :
(i) a2 + b2 = (a + b)2 - 2ab,
if a + b and ab are given
(ii) a 2 + b2 = (a - b)2 + 2ab
if a - b and ab are given
(iii) a + b =
(a − b )2 + 4ab
if a - b and ab are given
(iv) a -b =
(a + b )2 − 4ab
if a + b and ab are given
2
(v) a 2 +
1 
= a +
a2 
1
 −2
a
(vi) a 2 +
1 
1
= a +  + 2
a
a2 
if a +
1
is given
a
if a -
1
is given
a
2
(vii) a 3 + b3 = (a + b)3 - 3ab(a + b)
3
3
3
(viii) a - b = (a - b) + 3ab(a - b)
Free
if (a + b) and ab are given
if (a - b) and ab are given
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3
(ix) x 3 +
1 
1

=  a +  − 3 a +
a
a3 

1

a
(x) a 3 −
1 
1
1

=  a −  + 3 a − 1  ,
a
a
a3 

if a +
3
1
is given
a
1

if  a −  is given
a

(xi) a4 + b4 = (a2 + b2)2 - 2a 2b2 = [(a + b)2 - 2ab]2 - 2a2b2, if (a + b) and ab are given
(xii) a4 - b4 = (a2 + b2) (a2 - b2) = [(a + b)2 -2ab](a + b) (a - b)
(xiii) a5 + b5 = (a3 + b3) (a2 + b2) - a2b2 (a + b)
ILLUSTRATION
Ex.1
Find the value of :
(i) 36x2 + 49y2 + 84xy, when x = 3, y = 6
(ii) 25x2 + 16y2 - 40xy, when x = 6, y = 7
Sol.
(i) 36x2 + 49y2 + 84xy
= (6x)2 + (7y)2 + 2 × (6x) × (7y)
= (6x + 7y)2
= (6 × 3 + 7 × 6)2[When x = 3, y = 6]
= (18 + 42)2
= (60)2
= 3600.
2
2
(ii) 25x + 16y - 40xy
2
Ans.
2
= (5x) + (4y) - 2 × (5x) × (4y)
= (5x - 4y)2
= (5 × 6 - 4 × 7)2 [When x = 6, y = 7]
= (30 - 28)2
= 22
=4
Ans.
Ex.2
If x2 +
Sol.
x2 +
1
1

= 23, find the value of  x +  .
x
x2


1
= 23
x2
⇒ x2 +
....(i)
1
+ 2 = 25
x2
[Adding 2 on both sides of (i)]
2
1
1
⇒ (x2) +   + 2 . x .
= 25
x
x
2
⇒
1

 x +  = (5)2
x

⇒ x+
Free
1
= 5 Ans.
x
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Ex.3
Prove that a2 + b2 + c2 - ab - bc - ca =
Sol.
Here, L.H.S.
1
(a − b )2 + (b − c )2 + (c − a )2 .
2
[
]
= a2 + b2 + c2 - ab -+ bc - ca
=
1
[2a2 + 2b2 + 2c2 - 2ab - 2bc - 2ca]
2
=
1 2
[(a - 2ab + b2) + (b2 - 2bc + c2) + (c2 - 2ca + a2)]
2
=
1
[(a - b)2 + (b - c)2 + (c - a)2]
2
= RHS
Ex.4
Evaluate :
(i) (107)2
Sol.
Hence Proved.
(i)
(ii) (94)2
(iii) (0.99)2
(107)2 = (100 + 7)2
= (100)2 + (7)2 + 2 × 100 × 7
= 10000 + 49 + 1400
Ans.
= 11449
(ii)
(94)
2
= (100 - 6)
2
= (100)2 + (6)2 - 2 × 100 × 6
= 10000 + 36 - 1200
Ans.
= 8836
(iii)
(0.99)
2
= (1 - 0.01)
2
= (1)2 + (0.01)2 - 2 × 1 × 0.01
= + 0.0001 - 0.02
Ans.
= 0.9801
NOTE : We may extend the formula for squaring a binomial to the squaring of a trinomial as given below.
(a + b + c)2 = [a + (b + c)]2
= a2 + (b + c)2 + 2 × a × (b + c)
2
[Using (b + c)2 = b2 + c2 + 2bc]
= a2 + b2 + c2 + 2bc + 2ab + 2ac
[Using the distributive law]
2
2
[Using the identity for the square of binomial]
= a + b + c + 2bc + 2 (b + c)
2
2
2
= a + b + c + 2ab + 2bc + 2ac
∴
2
(a + b + c) = a2 + b2 + c2 + 2ab + 2bc + 2ac
Ex.5
Simplify : (3x + 4)3 - (3x - 4)3.
Sol.
We have,
(3x + 4)3 - (3x - 4)3
= [(3x)3 + (4)3 + 3 × 3x × 4 × (3x + 4)] - [(3x)3 - (4)3 - 3 × 3x × 4 × (3x - 4)]
= [273 + 64 + 36x (3x + 4)] - [273 - 64 - 36x (3x - 4)]
= [27x3 + 64 + 108x2 + 144x] - [27x3 - 64 - 108x2 + 144x]
= 27x3 + 64 + 108x2 + 144x - 27x3 + 64 + 108x2 - 144x
= 128 + 216x2
∵ (3x + 4)3 - (3x - 4)3 = 128 + 216x2
Free
Ans.
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Ex.6
Evaluate :
(i) (1005)3
Sol.
(i)
(ii) (997)3
(1005)3 = (1000 + 5)3
= (1000)3 + (5)3 + 3 × 1000 × 5 × (1000 + 5)
= 1000000000 + 125 + 15000 + (1000 + 5)
= 1000000000 + 125 + 15000000 + 75000
Ans.
= 1015075125.
(ii)
(997)3 = (1000 - 3)3
= (1000)3 - (3)3 - 3 × 1000 × 3 × (1000 - 3)
= 1000000000 - 27 - 9000 × (1000 - 3)
= 1000000000 - 27 - 900000 + 27000
Ans.
= 991026973
1
1
= 5, find the value of x3 - 3
x
x
Ex.7
If x -
Sol.
We have, x −
Ex.8
1
=5
x
...(i)
⇒


13
x −
= ( 5)3 

x


⇒
x3 −
1
1
1
− 3x. . x −  = 125
x
x
x3
⇒
x3 −
1
1

− 3 x −  = 125
x
x3


⇒
x3 −
1
− 3 × 5 = 125
x3
⇒
x3 −
1
− 15 = 125
x3
⇒
x3 −
1
= ( 125 + 15) = 140
x3
[Cubing both sides of (i)]
1

[Substituting  x −  = 5]
x

Ans.
Find the following products of the following expression :
(i) (4x + 3y) (16x2 - 12xy + 9y2)
Sol.
(i)
2
(ii) (5x - 2y) (25x2 + 10xy + 4y2)
2
(4x + 3y) (16x - 12xy + 9y )
= (4x + 3y) [(4x)2 - (4x) × (3y) + (3y)2]
= (x + b) (x2 - ab + b2)
[Where a = 4x, b = 3y]
= a3 + b3
= (4x)3 + (3y)3 = 64x3 + 27y3
(ii)
2
Ans.
2
(5x - 2y) (25x + 10xy + 4y )
= (5x - 2y) [(5x2 + (5x) × (2y) + (2y)2]
= (a - b) (a2 + ab + b2)
3
=a -b
[Where a = 5x, b = 2y]
3
= (5x)3 - (2y)3
= 125x3 - 8y3
Free
Ans.
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(a
2
) (
Simplify :
Sol.
Here a 2 − b 2 + b 2 − c 2
(
(a
∴
2
− b2
) (
3
)
3
3
− b2 + v2 − c2 + c2 − a2
.
(a − b)3 + (b − c )3 + (c − a )3
Ex.9
) (
) + (b
3
2
− c2
) + (c
) + (c
)
)
3
2
− a2 = 0
3
2
− a2
3
(
)(
)(
= 3 a2 − b2 b2 − c2 c2 − a2
)
(a − b) + (b − c ) + (c − a ) = 0
Also,
∴
(a − b)3 + (b − c )3 + (c − a )3 = 3(a − b)(b − c )(c − a )
∴
Given expression
=
3(a − b )(a + b )(b − c )(b + c )(c − a )(c + a )
3(a − b )(b − c )(c − a )
=
3(a − b )(a + b )(b − c )(b + c )(c − a )(c + a )
3(a − b )(b − c )(c − a )
= (a + b )(b + c)(c + a )
Ans.
Ex.10
Prove that : (x - y)3 + (y - z)3 + (z - x)3 = 3(x - y) (y - z) (z - x).
Sol.
Let (x - y) = a, (y - z) = b and (z - x) = c.
Then,
3
a + b + c = (x - y) + (y - z) + (z - x) = 0
3
∴ a + b + c3 = 3abc
Or (x - y)3 + (y - z)3 + (z - x)3 = 3(x - y) (y - z) (z - x)
Ex.11
Find the value of (28)3 - (78)3 + (50)3.
Sol.
Let a = 28, b = - 78, c = 50
Then,
∴
Ans.
a + b + c = 28 - 78 + 50 = 0
3
a + b3 + c3 = 3abc.
So, (28)3 + (-78)3 + (50)3 = 3 × 28 × (-78) × 50
Ex.12
If a + b + c = 9 and ab + bc + ac = 26, find the value of a3 + b3 + c3 - 3abc.
Sol.
We have a + b + c = 9
...(i)
⇒ (a + b + c)2 = 81
[On squaring both sides of (i)]
⇒ a + b + c + 2(ab + bc + ac) = 81
2
2
2
⇒ a 2 + b2 + c2 + 2 × 26 = 81
[ ∵ ab + bc + ac = 26]
⇒ a 2 + b2 + c2 = (81 - 52)
⇒ a 2 + b2 + 2 = 29.
Now, we have
a 3 + b3 + c3 - 3abc
= (a + b + c) (a 2 + b2 + c2 - ab - bc - ac)
= (a + b + c) [(a2 + b2 + c2) - (ab + bc + ac)]
= 9 × [(29 - 26)]
= (9 × 3)
= 27
Free
Ans.
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(b) A Special Product :
We have (x + a) (x + b) = x (x + b) + a (x + b)
= x2 + xb + ax + ab
= x2 + bx + ax + ab
= x2 + ax + bx + ab
= x2 + (a + b) x + ab
Thus, we have the following identity
(x + a) (x + b) = x2 + (a + b)x + ab.
Ex.13
Find the following products :
(i) (x + 2) (x + 3)
(ii) (y - 4) (y + 3)
(v) (2x - 3) (2x - 5)
[ ∵ xb = bx]
(ii) (x + 7) (x - 2)
(iv) (y - 7) (y + 3)
(vi) (3x + 4) (3x - 5)
Using the identity : (x + a) (x + b) = x2 + (a + b) x + ab, we have
Sol.
(i) (x + 2) (x + 3) = x2 + (2 + 3) x + 2 × 3
= x2 + 5x + 6.
Ans.
(ii) (x + 7) (x - 2) = (x + 7) (x + (-2))
= x2 + 7x + (-2)x + 7 × (-2)
= x2 + 5x - 14.
Ans.
(iii) (y - 4) (y- 3) = {y + (-4)} {y+ (-3)}
= y2 + {(-4) + (-3)}y + (-4) × (-3)
Ans.
= y2 - 7y + 12
(iv) (y - 7) (y + 3) = {y + (-7)} (y + 3)
= y2 + {(-7) + 3} + (-7) × 3
= y2 - 4y - 21.
Ans.
(v) (2x - 3) (2x + 5) = (y - 3) (y + 5), where y = 2x
= {y + (-3)} (y + 5)
= y2 + {(-3) + 5} y + (-3) × 5
= y2 + 2y - 15
= (2x)2 + 2 × 2x - 15
= 4x2 + 4x - 15.
Ans.
(vi) (3x + 4) (3x - 5) = (y + 4) (y - 5), where y = 3x
= (y+ 4) {y + (-5)}
= y2 + {4 + (-5)} + 4 × (-5)
= y2 - y - 20
= (3x)2 - 3x - 20
= 9x2 - 3x - 20.
Ex.14
Sol.
= 1600 - 320 + 15
= 1615 - 320
= 1295
Free
Ans.
Evaluate : (i) 35 × 37
(ii) 103 × 96
(i) 35 × 37 = (40 - 5) (40 - 3)
= (40 + (-5)) (40 + (-3))
= 402 + (-5 - 3) × 40 + (-5 × - 3)
Ans.
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(ii)
103 × 96
= (100 + 3) [100 + (-4)]
= 1002 + (3 + (-4)) × 100 + (3 × - 4)
= 10000 - 100 - 12
Ans.
= 9888
FACTORS OF A POLYNOMIAL
If a polynomial f(x) can be written as a product of two or more other polynomial f1(x), f2(x), f3(x),..... then
each of the polynomials f1(x), f2(x),..... is called a factor of polynomial f(x). The method of finding the factors
of a polynomials is called factorisations.
(a) Factorisation by Making a Trinomial a Perfect Square :
Ex.15
81a2b2c2 + 64a6b2 - 144a 4b2c
Sol.
81a2b2bc2 + 64a6b2 - 144a4b2c
= [9abc]2 -2 [9abc][8a3b] + [8a3b]2
= [9abc - 8a 3b]2 = a 2b2[9c - 8a2]2 Ans.
2
Ex.16
1 
1
1
1




 3a −  −  3a −  + 9 +  c + − 2a  3a − − 3 
b 
b
b
b




Sol.
1
1
1
1





 3a −  − 6 3a −  + 9 +  c + − 2a  3a − − 3 
b
b
b
b





2
2
1
1
1
1





=  3a −  − 2.3 3a −  + ( 3)2 +  c + − 2a  3a − − 3 
b
b
b
b





2
1
1
1

 


=  3a − − 3  +  c + − 2 a  3a − − 3 
b
b
b

 


1
1
1



=  3a − − 3  3a − + 3 + − 2 a
b
b
b



1


=  3a − − 3 [a + c − 3]
b


Ans.
(b) Factorisation by Using the Formula for the Difference of Two Squares :
a2 - b2 = ( + b) (a - b)
Ex.17
Sol.
Factorise
:
4(2a + 3b - 4c)2 - (a - 4b + 5c).2
= 4(2a + 3b - 4c)2 - (a - 4b + 5c)2
= [2(2a + 3b - 4c)]2 - (a - 4b + 5c)2
= [4a + 6b - 8c + a - 4b + 5c] [4a + 6b - 8c - a + 4b - 5c]
= [5a + 2b - 3c] [3a + 10b - 13c]
Free
Ans.
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Ex.18
Factorise : 4x 2 +
Sol.
4x2 +
1
+ 2 − 9y2 .
4x2
1
+ 2 − 9y2
4x2
2
 1   1 
= ( 2 x)2 + 2.( 2 x ).  +   − ( 3 y )2
 2x   2x 
2
1 

2
=  2x +
 − ( 3y )
2x 

1
1



=  2x +
+ 3 y  2 x +
− 3y 
2x
2x



Ans.
1
− 3.
a4
Ex.19
Factorise : x 4 +
Sol.
 1 
 1 
( a 2 )2 +  2  − 2.( a 2 ) 2  − 1
a 
a 
2
2
1 

=  a 2 − 2  − ( 1) 2
a 

1
1



=  a 2 − 2 + 1  a 2 − 2 − 1 
a
a



Ans.
Ex.20
Factorise : x4 + x2y2 + y4.
Sol.
x2 + x2y2 + y4 = (x2)2 + 2.x2.y2 + (y2)2 - x2y2
= x2 + y2)2 - (xy)2
= (x2 + y2 + xy( (x2 + y2 - xy)
Ans.
(c) Factorisation by Using Formula of a3 + b3 and a3 - b3 :
Ex.21
Factorise : 64a13b + 343ab13.
Sol.
64a13b + 343ab13 = ab[64a12 + 343b12]
= ab[(4a4)3 + (7b4)3]
= ab[4a4 + 7b4] [(4a 4)2 - (4a 4) (7b4) + (7b4)2]
= ab[4a4 + 7b4][16a8 - 28a 4b4 + 49b8]
Ans.
x
- q2r3x
q
Ex.22
Factorise : p 3q2x4 + 3p 2qx3 + 3px2 +
Sol.
In above question, If we take common then it may become in the form of 3 + b3.
∴
Free
p 3q2x4 + 3p2qx3 + 3px2 +
x
- q2r3x
q
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=
x 3 3 3
[p q x + 3p 2q2x2 + 3pqx + 1 - q3r3]
q
=
x
[(pqx)3 + 3(pqx)2 .1 + 3pqx . (1)2 + (1)3 - q3r3]
q
Let pqx = A & 1 = B
=
x
[A3 + 3A2B + 3AB2 + B3 - q3r3]
q
=
x
x
[(pqx + 1)3 - (qr)3] = [pqx + 1 - qr][(pqx + 1)2 + (pqx + 1) qr + (qr)2]
q
q
=
x
[pqx + 1 - qr][p2q2x2 + 1 + 2pqx + pq2xr + qr + q2r2] Ans.
q
Ex.23
Factories : x3 - 6x2 + 32
Sol :
x3 + 32 - 6x2
= x3 + 8 + 24 - 6x2
= [(x)3 + (2)3] + 6[4 - x2]
= (x + 2)[x2 - 2x + 4] + 6[2 + x][2 - x]
= (x + 2) [x2 -2x + 4 + 6(2 - x)]
= (x + 2)[x2 -2x + 4 + 12 - 6x]
= (x + 2) [x2 - 8x + 16]
= (x + 2) (x - 4)2 Ans.
Free
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EXERCISE
OBJECTIVE DPP - 4.1
1.
The product of (x + a) (x + b) is :
(A) x2 + (a + b) x + ab
2.
4.
5.
(D) 10470
(C) x2 + y2 + z2 + 2xy - 2yz - 2zx
(D) x2 + y2 - z2 + 2zy - 2yz - 2zx
The value of (x + 2y + 2z)2 + (x - 2y - 2z)2 is:
(A) 2x2 + 8y2 + 8z2
(B) 2x2 + 8y2 + 8z2 + 8xyz
(C) 2x2 + 8y2 + 8z 2 - 8yz
(D) 2x2 + 8y2 + 8z 2 + 16yz
The value of 25x2 + 16y2 + 40 xy at x = 1 and y = -1 is :
(B) -49
(C) 1
(D) None of these
On simplifying (a + b)3 + (a - b)3 + 6a(a2 - b2) we get :
Find the value of
(B) 8a2b
(C) 8a 3b
(D) 8a 3
a 3 + b 3 + c 3 − 3abc
, when a = -5, 5 = -6 , c = 10.
ab + bc + ca − a 2 − b 2 − c 2
(B) -1
(C) 2
(D) -2
If (x + y + z) = 1, xy + yz + zx = -1 xyz = -1 then value of x3 + y3 + z 3 is :
(A) -1
9.
(C) 14700
(B) x2 + y2 - z2 - 2xy + yz + 2zx
(A) 1
8.
(B) 14800
(A) x2 + y2 + z2 + 2xy + 2yz + 2zx
(A) 8a2
7.
(D) x2 + (a - b)x - ab.
The expansion of (x + y - z)2 is :
(A) 81
6.
(C) a2 + (a - b)x + ab
The value of 150 × 98 is :
(A) 10047
3
(B) x2 - (a - b) x + ab
(B) 1
(C) 2
(D) -2
In method of factorisation of an algebraic expression. Which of the following statement is false ?
(A) Taking out a common factor from two or more terms.
(B) Taking out a common factor from a group of terms.
(C) By using remainder theorem.
(D) By using standard identities.
10.
Factors of (a + b)3 - (a - b)3 is :
(A) 2ab(3a 2 + b2)
11.
(C) 2b(3a2 + b2)
(D) 3a 2 + b2
(C) Both 0 & 1
(D) Not defined
Degree of zero polynomial is :
(A) 0
Free
(B) ab(3a 2 + b2)
(B) 1
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SUBJECTIVE DPP 4.2
1
1
= 119 , then find the value of − 3 .
a4
a
1.
If a 4 +
2.
If x = 152, y = -91 find the value of 9x2 + 30xy + 25y2.
3.
Evaluate :
(i) (5x + 4y)2
1

(iii)  2 x − 
x

(ii) (4x - 5y)2
4.
If x + y = 3 and xy = - 18, find the value of x3 + y3.
5.
If x 2 +
6.
Evaluate :
1
1
= 51 find the value of x 3 − 3 . .
2
x
x
3
3
(iii) (0.2)3 - (0.3)3 + (0.1)3
Find the product of :
1
(ii) ( x + )( x + 5)
5
(i) (x + 4) (x + 7)
8.
3
 1 1 5
(ii)   +   −  
2 3 6
(i) 253 - 753 + 503
7.
2
1
(iii) (P2 + 16) (P 2 − )
4
Evaluate :
(i) 102 × 106
(ii) 994 × 1006
(iii) 34 × 36
9.
Factorise : 4x4 + (7a)4.
10.
Factorise : x12 = 1.
11.
Evaluate
12.
Write the following polynomials in standard forms :
(b − c )2
(c − a )2
( a − b )2
.
+
+
(b − c )(c − a ) (a − b )(c − a ) (a − b )(b − c )
(i) x6 - 3a 4 +
2 x + 5x2 + 7x5 + 4
(ii) m7 + 8m5 + 4m6 + 6m - 3m2 - 11
13.
Factorise : (x + 1) (x + 2) (x + 3) (x + 4) - 3.
14.
Factorise : 64a3 - 27b3 - 144a2b + 108ab2.
15.
Factorise : x4 + 2x3y - 2xy3 - y4.
Free
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